SOLUTION: The roots z1,z2 Nd z3 of the equation x^3+3ax^2+3bx+c=0 in which a,b and c are complex numbers, correspond to the points A,B and C on the Argand plane. Find the C.G. of ∆ABC

C.G. of a triangle is the center of gravity of the triangle.


The center of gravity of the triangle with the vertices A, B and C is the point with coordinates


       and    .


Geometrically, this point is nothing else as   .


According to Vieta's theorem,  the sum    is the coefficient at  x^2  of the given equation  with the opposite sign, i.e.  -3a.


Thus the center of gravity of the given triangle is the complex number  -a.

If  a^2 = b,  then  the given equation


    x^3 + 3ax^2 + 3bx + c = 0


becomes


    x^3 + 3ax^2 + 3a^2*x + c = 0,    or, equivalently,


    (x^3 + 3ax^3 + 3a^2*x + a^3) + (c-a^3) = 0,   

    (x+a)^3 = -(c-a^3),

    x + a = ,


    x = - a + .     (1)


If you are familiar with the basics of complex number theory, you will recognize that the formula (1)

describes the three complex numbers around the point "-a"  (the center of gravity of the triangle ABC) turned at the angle 120 degrees

one relative the other.


So the triangle ABC is an equilateral triangle.